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So you have 2 conditions to have a removable discontinuity. The first one is simply that the value must not be defined at that point, which is pretty obvious that is isn't because of the denominator. The second condition is basically the difference between whether or not you have an asymptote at your undefined point or simply a hole. The way to tell whether or not its an asymptote or a whole is if you factored your function, would the factor that is undefined cancel or remain? If the factor on bottom cancels, it is merely a hole, if it does not cancel then it is an asymptote and CANNOT be removed. So I would factor your numerator and see if the denominator cancels. If it does, then simply plug in 4 into your result and you'll get the value you're looking for.
can we work on this together?
i've tried doing this on my own but seem to be getting something wrong.
Sure. So I would factor the numerator and see if x-4 cancels. If it cancels, you have a hole, if it does not cancel, you have an asymptote and it cannot be removed. You know how to factor the numerator, right?
let me try.
i got 2(x^2+3x-25) @Psymon
-28you mean? And after that, you still have to further factor the x^2 + 3x - 28.
Find factors of -28 that will add up to positive 3.
Right. So When you further factor the numerator, you should have 2(x±__)(x±__). Its just remember how to factor that x^2 + 3x - 28 portion.
Exactly. In turn, that cancels out the denominator, which means the discontinuity is removable and not just an asymptote. Not only that, since the term that would make your function undefined disappeared, you can simply plug in 4 to find out the value you need.
we're left with 2 (x+7)
so we plug in 4?
That's all ya do.
does this apply to every problem with continuity?
Yes. Most of the work is finding a way to eliminate the portion that would give you an undefined answer. Once you manage that, you just plug in the value that the limit is approaching. Of course there are times when you cannot remove the undefined portion, in which that means the limit does not exist and you are merely approaching an asymptote.
ohhh can i practice a little, then ask you for a tricky problem like the one you just mentioned? it would really help.
That works, I have no problem with that.
what about a JUMP DISCONTINUITY? same thing?
That's new terminology for me. I probably know what it is but have not heard it said that way. Could you give me an example?
Ah, just one sided limits. We never really gave a name for it before, just that it had to do with one-sided, lol.
oh so how do we begin with this one? sketch a graph?
i can do graphs but i'd like to do it a simpler way since i won't have enough time when it comes to a final
I apologize, I guess I was making sure what jump kind of meant since I did not have a picture. But yes, all you really need to confirm is that when x = 6 on both functions that they have different limits. It is pretty obvious that they do. As far as the left and right part goes, you need to see which function to just check and see which function is approaching from the left and which one is approaching from the right. Since 8x-8 < 6, that must be the function approaching from the left. Therefore, f(6) of 8x-8 is the value you will use for limit from the left. Since the other function is the one that will becoming from the right, it's value at x = 6 will be the value you use for the right-sided limit. Basically, you just need to see which function to plug the value into.
The delay was graphing and looking at so I knew exactly what to say and to get my brain to work, lol.
it's okay, so how can we work this prob out?
|dw:1375046904734:dw| I know you know how to graph, but I wanted the visual up before we move on. So you just want to see which function is from the left and which one is from the right. When it gives you a piece-wise function, it's pretty obvious which one is going to be coming from the left and which one from the right. Once you determine that, just plug in x = 6 into the graph that comes from the left and use that value for lim ->6 - Once you do that, plug x = 6 into the other function, since you know it's the one on the right, and find lim x-> 6 +
A lot of the time you may have to determine the behavior on your own, but the piece-wise just gave you the behavior so you didn't have to worry about it. From there it was just picking the correct function for left and for right.
In regards to removable or non-removable,it works the same way. if one of the functions has an asymptote at the jump point, then one of the sides or both may have a limit that DNE on that side, left or right. Maybe another example?
so to this problem we HAVE to sketch only?
In this case, no, because the piece-wise function tells us which function to use on the left and on the right. The sketching would only come in if you had maybe one function that wasn't a piece-wise. Like if the function was 1/[(2x^2 -3] For something like that you would have to test points on the left and right and see if they're going up to infinity, down to negative infinity, etc. Any of these piece-wise functions won't require s ketch, I just wanted to draw visual aid.
Holy cow my typing and grammar are horrible today o.o
Maybe if you have another jump problem we can look at it.
I'm going to need to head out, I apologize. Good luck with the limits.